$|\alpha _{n}| \le c_n$, then $\sum{\alpha _{n}}$ converges absolutely

Question

Can you help me understand the part of the proof. Thanks

Let $\sum{c_n}$ be a series of real numbers $\ge 0$ which converges. If $|\alpha _{n}| \le c_n$ for all n, then the series $\sum{\alpha _{n}}$ converges absolutely.

The proof says: "The partial sums $\sum_{k=1}^{n} {c_k}$" are bounded by assumption (I understand this, since it is convergent), whence the partial sums $\sum_{k=1}^{n}|{\alpha_k}| \le \sum_{k=1}^{n}{c_k}$ are also bounded, and the absolute convergence follows.

I don't understand why it follows from the boundedness

Answer 1

A bounded monotone sequence is convergent.

Answer 2

Since $|a_k| \ge 0$ for all $k$, the sequence $S_N :=\sum_{k=1}^N|a_k|$ is a monotonically increasing series. To show the convergence of $S_N$ we just need to know that $S_N$ is bounded above. This is where the hypothesis $|a_k|\le c_k$ is needed.

Defining $T_N:=\sum_{k=1}^N c_k$ we know that $T_N$ converges by hypothesis. Let us call $T$ the limit of the sequence $(T_N)$, which exists by hypothesis.

$|a_1|\le c_1$ so $S_1\le T_1$ and by induction it can be shown that $S_n\le T_n\le T$ for all $n\in \mathbb{N}$.

Therefore $(S_N)$ is bounded above by $T$ and so we can conclude that $S_N$ is convergent.